Acceleration due to gravity at the given height, $g\text{'}=\frac{GM}{{\left(3R\right)}^2}=\frac{1}{9}\frac{GM}{{\left(R\right)}^2}=\frac{1}{9}g$, where $g$ is acceleration due to gravity at the surface of the Earth.

Now, time period is given by, $T=2\pi \sqrt{\frac{\mathcal{l}}{g\text{'}}}$

Also the time period of second pendulum is $2$ sec.

Therefore, $T=2 \sec$

Then, $2=2\pi \sqrt{\frac{\mathcal{l}}{g}\times 9}$

$\Rightarrow \mathcal{l}=\frac{1}{9} \mathrm{m}$